# a compact metric space is second countable

###### Proof.

Let $(X,d)$ be a compact  metric space, and for each $n\in\mathbb{Z}^{+}$ define $\mathcal{A}_{n}=\{B(x,1/n):x\in X\}$, where $B(x,1/n)$ denotes the open ball centered about $x$ of http://planetmath.org/node/1296radius $1/n$. Each such collection  is an open cover of the compact space $X$, so for each $n\in\mathbb{Z}^{+}$ there exists a finite collection $\mathcal{B}_{n}\subseteq\mathcal{A}_{n}$ that $X$. Put $\mathcal{B}=\bigcup_{n=1}^{\infty}\mathcal{B}_{n}$. Being a countable  union of finite sets  , it follows that $\mathcal{B}$ is countable; we assert that it forms a basis for the metric topology on $X$. The first property of a basis is satisfied trivially, as each set $\mathcal{B}_{n}$ is an open cover of $X$. For the second property, let $x,x_{1},x_{2}\in X$, $n_{1},n_{2}\in\mathbb{Z}^{+}$, and suppose $x\in B(x_{1},1/n_{1})\cap B(x_{2},1/n_{2})$. Because the sets $B(x_{1},1/n_{1})$ and $B(x_{2},1/n_{2})$ are open in the metric topology on $X$, their intersection  is also open, so there exists $\epsilon>0$ such that $B(x,\epsilon)\subseteq B(x_{1},1/n_{1})\cap B(x_{2},1/n_{2})$. Select $N\in\mathbb{Z}^{+}$ such that $1/N<\epsilon$. There must exist $x_{3}\in X$ such that $x\in B(x_{3},1/2N)$ (since $\mathcal{B}_{2N}$ is an open cover of $X$). To see that $B(x_{3},1/2N)\subseteq B(x_{1},1/n_{1})\cap B(x_{2},1/n_{2})$, let $y\in B(x_{3},1/2N)$. Then we have

 $d(x,y)\leq d(x,x_{3})+d(x_{3},y)<\dfrac{1}{2N}+\dfrac{1}{2N}=\dfrac{1}{N}<% \epsilon\text{,}$ (1)

so that $y\in B(x,\epsilon)$, from which it follows that $y\in B(x_{1},1/n_{1})\cap B(x_{2},1/n_{2})$, hence that $B(x_{3},1/2N)\subseteq B(x_{1},1/n_{1})\cap B(x_{2},1/n_{2})$. Thus the countable collection $\mathcal{B}$ forms a basis for a topology  on $X$; the verification that the topology by $\mathcal{B}$ is in fact the metric topology follows by an to that used to verify the second property of a basis, and completes      the proof that $X$ is second countable. ∎

It is worth nothing that, because a countable union of countable sets is countable, it would have been sufficient to assume that $(X,d)$ was a Lindelöf space.

 Title a compact metric space is second countable Canonical name ACompactMetricSpaceIsSecondCountable Date of creation 2013-03-22 17:00:49 Last modified on 2013-03-22 17:00:49 Owner azdbacks4234 (14155) Last modified by azdbacks4234 (14155) Numerical id 17 Author azdbacks4234 (14155) Entry type Theorem Classification msc 54D70 Related topic MetricSpace Related topic Compact Related topic Lindelof Related topic Ball Related topic basisTopologicalSpace Related topic Cover Related topic BasisTopologicalSpace